Question

Let \(p_1, p_2, p_3\) be the altitudes of a triangle ABC drawn through the vertices A, B, C respectively. If \(r_1=4, r_2=6, r_3=12\) are the ex-radii of triangle ABC then \( \frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} = \)

• A. \( \frac{25}{72} \)
• B. \( \frac{25}{144} \)
• C. \( \frac{25}{288} \)
• D. \( \frac{25}{216} \)

Hint:

For problems involving radii (in-radius, ex-radii) and other triangle properties (altitudes, sides, area), remember these key relations: \( \frac{1}{r} = \sum \frac{1}{r_i} \), \( \Delta^2 = r r_1 r_2 r_3 \), \( \Delta = rs \), and \( r_1 = \Delta/(s-a) \). These allow you to find all properties of the triangle from a few given values.

Answer 1

The Correct Option is C

We use the formula relating the altitude to a side and the area of the triangle (\(\Delta\)): \( p_1 = \frac{2\Delta}{a} \), so \( \frac{1}{p_1} = \frac{a}{2\Delta} \).
The expression becomes \( \frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} = \frac{a^2}{(2\Delta)^2} + \frac{b^2}{(2\Delta)^2} + \frac{c^2}{(2\Delta)^2} = \frac{a^2+b^2+c^2}{4\Delta^2} \).
We can find the properties of the triangle from the ex-radii. First, find the in-radius \(r\).
The relation is \( \frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} \).
\( \frac{1}{r} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12} = \frac{3+2+1}{12} = \frac{6}{12} = \frac{1}{2} \). So, \( r=2 \).
We also know \( \Delta^2 = r r_1 r_2 r_3 = 2 \cdot 4 \cdot 6 \cdot 12 = 576 \). So, \( \Delta = 24 \).
Using \( r = \Delta/s \), we find the semi-perimeter \( s = \Delta/r = 24/2 = 12 \).
Now we find the sides using \( r_1 = \Delta/(s-a) \), etc.
\( 4 = 24/(12-a) \implies 12-a=6 \implies a=6 \).
\( 6 = 24/(12-b) \implies 12-b=4 \implies b=8 \).
\( 12 = 24/(12-c) \implies 12-c=2 \implies c=10 \).
Now we calculate \( a^2+b^2+c^2 = 6^2+8^2+10^2 = 36+64+100 = 200 \).
Finally, substitute the values into our expression.
\( \frac{a^2+b^2+c^2}{4\Delta^2} = \frac{200}{4(576)} = \frac{50}{576} = \frac{25}{288} \).